Fixed Point Iteration

Copyright (C) 2020 Andreas Kloeckner

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In [1]:
import numpy as np

import matplotlib.pyplot as pt

Task: Find a root of the function below by fixed point iteration.

In [2]:
x = np.linspace(0, 4.5, 200)



def f(x):

    return x**2 - x - 2



pt.plot(x, f(x))

pt.grid()

Actual roots: $2$ and $-1$. Here: focusing on $x=2$.

We can choose a wide variety of functions that have a fixed point at the root $x=2$:

(These are chosen knowing the root. But here we are only out to study the behavior of fixed point iteration, not the finding of fixed point functions--so that is OK.)

In [3]:
def fp1(x): return x**2-2

def fp2(x): return np.sqrt(x+2)

def fp3(x): return 1+2/x

def fp4(x): return (x**2+2)/(2*x-1)



fixed_point_functions = [fp1, fp2, fp3, fp4]
In [4]:
for fp in fixed_point_functions:

    pt.plot(x, fp(x), label=fp.__name__)

pt.ylim([0, 3])

pt.legend(loc="best")

<ipython-input-3-8e76f485a071>:3: RuntimeWarning: divide by zero encountered in true_divide
def fp3(x): return 1+2/x
Out[4]:
<matplotlib.legend.Legend at 0x7efd766b6df0>

Common feature?

In [5]:
#clear

for fp in fixed_point_functions:

    print(fp(2))

    

# All functions have 2 as a fixed point.

2
2.0
2.0
2.0
In [7]:
z = 2.1; fp = fp1

#z = 1; fp = fp2

#z = 1; fp = fp3

#z = 1; fp = fp4



n_iterations = 4



pt.figure(figsize=(8,8))

pt.plot(x, fp(x), label=fp.__name__)

pt.plot(x, x, "--", label="$y=x$")

pt.gca().set_aspect("equal")

pt.xlim([0, 4])

pt.ylim([-0.5, 4])

pt.legend(loc="best")



for i in range(n_iterations):

    z_new = fp(z)

    

    pt.arrow(z, z, 0, z_new-z)

    pt.arrow(z, z_new, z_new-z, 0)

    

    z = z_new

    print(z)

2.41
3.8081000000000005
12.501625610000003
154.29064289260796
In [ ]: